0 \\ \\ y_1= \frac{-b- \sqrt{D} }{2a} = \frac{15- \sqrt{9} }{2*1} = \frac{12}{2} =6 \\ \\ y_2= \frac{-b+ \sqrt{D} }{2a} = \frac{15+ \sqrt{9} }{2*1} = \frac{18}{2} = 9 \\ \\ x_1 = 15-6=9 \\ \\ x_2 = 15 - 9=6
" alt="\left \{ {{x+y=15} \atop {x*y=-54}} \right. \\ \\ x=15-y \\ \\ (15-y)*y-54=0 \\ \\ y^2-15y+54=0 \\ \\ D=b^2-4ac=15^2-4*1*54=9>0 \\ \\ y_1= \frac{-b- \sqrt{D} }{2a} = \frac{15- \sqrt{9} }{2*1} = \frac{12}{2} =6 \\ \\ y_2= \frac{-b+ \sqrt{D} }{2a} = \frac{15+ \sqrt{9} }{2*1} = \frac{18}{2} = 9 \\ \\ x_1 = 15-6=9 \\ \\ x_2 = 15 - 9=6
" align="absmiddle" class="latex-formula">
или по теореме Виета y₁ + y₂ = 15 y₁ * y₂ = 54
y₁ = 6
y₂ = 9
x₁ = 15 - 6 = 9
x₂ = 15 - 9 = 6
Ответ:
D. 6; 9